−GQ dz;
a positive quantity if dz is negative, and vice versa. The resultant pressure parallel to the axis of the pipe is p − (p + dp) = −dp ℔ per square foot of the cross section. The work of this pressure on the volume Q is
−Q dp.
The only remaining force doing work on the system is the friction against the surface of the pipe. The area of that surface is χdl.
The work expended in overcoming the frictional resistance per second is (see § 66, eq. 3)
−ζGχ dl v3/2g,
or, since Q = Ωv,
−ζG (χ/Ω) Q (v2/2g) dl;
the negative sign being taken because the work is done against a resistance. Adding all these portions of work, and equating the result to zero, since the motion is uniform,—